[tex]\dfrac{\mathrm dy}{\mathrm dx}=2\sin(4x-2y)[/tex]
Not sure what method you're referring to... but one substitution comes to mind: let [tex]v(x)=4x-2y(x)[/tex], so that [tex]\frac{\mathrm dv}{\mathrm dx}=4-2\frac{\mathrm dy}{\mathrm dx}[/tex]. Then
[tex]\dfrac12\left(4-\dfrac{\mathrm dv}{\mathrm dx}\right)=2\sin v\implies\dfrac{\mathrm dv}{\mathrm dx}=4(1-\sin v)[/tex]
This ODE is separable, as
[tex]\dfrac{\mathrm dv}{1-\sin v}=4\,\mathrm dx[/tex]
Integrate both sides; on the left, we have
[tex]\dfrac1{1-\sin v}=\dfrac{1+\sin v}{1-\sin^2v}=\dfrac{1+\sin v}{\cos^2v}=\sec^2v+\sec v\tan v[/tex]
which has a recognizable antiderivative, giving us
[tex]\tan v+\sec v=4x+C[/tex]
[tex]\implies\tan(4x-2y)+\sec(4x-2y)=4x+C[/tex]
so that the solution is
[tex]F(x,y)=\boxed{\tan(4x-2y)+\sec(4x-2y)-4x=C}[/tex]